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This is a question based on Berkson's Fallacy. Is the following inequality true? If so, how to prove?

$ P(A | A \cup B ) \geq P(A) $

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    $\begingroup$ The inequality needs quantifiers. Do you mean for all $A$ and $B$ or there can exist $A$ and $B$? It also needs to be stated a little more precisely, because obviously $P(A|A\cup B)=P(A)$ violates your strict inequality whenever $A\cup B$ is the entire space. $\endgroup$ – whuber Jul 9 '15 at 20:02
  • $\begingroup$ In the edited version (with the greater-than sign), it should be true for all $A$ and $B$, right? Isn't that the nature of the "paradox"? $\endgroup$ – Matt Krause Jul 10 '15 at 2:50
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$$ \begin{align} P( A \mid A \cup B) &= \frac{P(A)}{P(A \cup B)} \\ &\geq P(A) . \end{align} $$

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    $\begingroup$ +1. Since possibly $P(A\cup B)=0$, this could be made rigorous by recasting it in the form $$P(A)=P(A|A\cup B)P(A\cup B) \le P(A|A\cup B)(1)=P(A|A\cup B).$$ The first equality is often taken as an axiom of probability while the second follows from the axiom that no probability, such as $P(A\cup B)$, can exceed $1$. $\endgroup$ – whuber Jul 10 '15 at 14:10
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I've mostly heard of it as Berkson's paradox and it refers to the spurious generation of associations when you are comparing an exposure and an outcome and you sample only individuals with either the exposure or the outcome. Suppose the population level association is:

$$ \begin{array}{c|ccc} & D & \bar{D} & \\ \hline E & n_{11} & n_{12} & n_{1.} \\ \bar{E} & n_{21} & n_{22} & n_{2.} \\ & n_{.1} & n_{.2} & \\ \end{array} $$

Then the relative risk for disease is given by:

$$ RR = \frac{n_{11} / n_{1.} }{n_{21} /n_{2.}}$$

However, in your sample you obtain the following:

$$ \begin{array}{c|ccc} & D & \bar{D} & \\ \hline E & n_{11} & n_{12} & n_{1.} \\ \bar{E} & n_{21} & 0 & n_{2.} - n{22}\\ & n_{.1} & n_{.2} - n{22} & \\ \end{array} $$

with the cell counts and margins proportional to the "population" above WLOG.

The estimated relative risk becomes:

$$ RR_{Berkson} = \frac{n_{11}/n_{1.}}{n_{21} / (n_{2.} - n_{22})} $$

which is biased except when $n_{22} = 0$.

In a less biostatistical fashion, assume $P(A \cup B ) \neq 1$ then $P(A | A \cup B) = \frac{P(A \cap A \cup B)}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)}$ and we're done by assumption.

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  • $\begingroup$ What is your interpretation of the question and your answer to it? It looks like it should be either "yes" or "no," but I don't see anything that definite in this answer. $\endgroup$ – whuber Jul 9 '15 at 20:03
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Yes, it is true that $P(A|A \cup B) \ge P(A)$.

I find that helps to think of the two quantities as fractions. Then the whole thing follows from these two facts:

  • The numerator of both probabilities consists of the number of $A$s present in the population. This is trivial in the case of the unconditional probability $P(A)$, and guaranteed by having the $A$ appear on both sides of the condition in the conditional probability $P(A|A \cup B)$. In other words, by construction, the conditional "restriction" can't throw out any $A$s.

  • The denominator of $P(A)$ is the sum of the number of $A$s in the population, plus the number of $B$s in it, plus the number of $C$s, $D$s, and whatever else happens to be in it. However, the denominator of $P(A|A \cup B)$ is merely the number of $A$s, plus the number of $B$s; the condition rules everything else out. Since these are all non-negative numbers, the denominator of $P(A)$ must be at least as large as the denominator of $P(A | A \cup B)$.

Thus, we have $$\begin{align} P(A) &= \frac{||A||}{||A|| + ||\textrm{ Everything else }||}\\ P(A|A\cup B) &=\frac{||A||}{||A|| + ||B|| - ||A \cap B||} \end{align}$$

Since the numerators are the same, but $P(A)$'s denominator is greater than or equal to $P(A|A \cup B)$'s, it must be true that $P(A|A \cup B) \ge P(A). \blacksquare$

People call this a paradox because it is true even if $A$ and $B$ are unrelated or even mutually exclusive. Suppose $P(A)$ is the probability that $A$ plays professional basketball and $B$ indicates that $B$ is terrible at basketball. The inequality still holds because the condition rules out people who are fair-to-middling at basketball (and don't happen to play for the Knicks), so the denominator is still smaller than the overall population.

More generally, it is true that the denominator of a conditional probabilty will always be less than or equal to the denominator an unconditional probability.

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  • $\begingroup$ Remember that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $A$ and $B$ may not be mutually exclusive. $\endgroup$ – dsaxton Jul 10 '15 at 17:37
  • $\begingroup$ @dsaxton, true--though that just shrinks the conditional's denominator even more... $\endgroup$ – Matt Krause Jul 10 '15 at 17:50
  • $\begingroup$ Is there a typo in the last sentence of your 2nd bullet? It contradicts the last sentence before the black square? $\endgroup$ – Greek - Area 51 Proposal Dec 25 '17 at 3:16

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